CPhill

tan (15°) =   [1 -cos 30 ] / sin 30  = [ 1 -sqrt (3)/2 ] / (1/2) ] =  [2 -sqrt (3)] / 1

sin (15°) =   [ 2 -sqrt 3 ] / sqrt [ 1^2  + (2 -sqrt 3)^2 ]   =  [2 -sqrt 3 ] /  sqrt [ 8 - 4sqrt 3]

cos (15°)  =  1 / sqrt [ 8 - 4sqrt 3]

tan (7.5°) =  [1 - cos 15 ] / sin 15  =  [ 1 - 1 / sqrt [8 - 4sqrt 3 ] ] / ( [ 2 -sqrt 3] / sqrt [ 8 -4sqrt 3] ) =

sqrt [ 8 -4sqrt 3 ] * [  1  - 1/ sqrt [ 8 -4sqrt 3] / [2 -sqrt 3]   =

(sqrt [ 8 - 4sqrt 3 ] [ sqrt [8 - 4sqrt 3 ] - 1 ] / sqrt [8 -4sqrt 3])  / [2 -sqrt 3]  =

(sqrt [ 8  -4sqrt 3 ] - 1)  /   [2 -sqrt 3 ]  =

[ sqrt [ 8 - 4sqrt 3] - 1 ] [ 2 + sqrt 3 ]  =    { sqrt [ 8 - 4sqrt 3 ]  = sqrt [( sqrt 6 - sqrt 2)^2] = sqrt 6 - sqrt 2 }

[(sqrt 6 - sqrt 2 )  -1 ] [ 2 + sqrt 3]  =

2sqrt 6 - 2sqrt 2 - 2 + sqrt 18 - sqrt 6 - sqrt 3   =

sqrt 6 - 2sqrt 2- 2 + 3sqrt 2 -sqrt 3 =

sqrt 6 + sqrt 2 - sqrt 3  - 2   =

sqrt (2)  + sqrt (6)  -sqrt (3)   - 2   =

sqrt (2) + sqrt (6) - sqrt (3)  - sqrt (4)

a + b =  8

c + d  =  7

C ( 19,7)  = 50388 ways

https://web2.0calc.com/questions/angle-problem_3

cos A =  24/ sqrt [ 24^2 + 7^2 ]  =  24  / 25

sin A = 7 / 25

tan (A/2)  =  [ 1 -cos A ] / sin A =  [ 1- 24/25 ] / [ 7/25 ] =  1 / 7

tan (A/2)  = CD / AC

(1/7)  = CD / 24

CD  = 24 / 7

Likewise

cos B = 7/25

sin B = 24/25

tan (B/2)  = [ 1 -cos B ] / sin B  =  [ 1 - 7/25 ] / [ 24/25]  =  18/24 = 3 / 4

tan (B/2)  = CE / BC

3/4  =  CE  / 7

CE =  21/4

Reflection over y = x just reverses the coordinates....so B = (7,5)

7! / ( 2!)  =  2520 identifiable sequences

(x-2) (x + 6) ≤  (x - 2) ( x + 5)

x^2 + 4x - 12 ≤ x^2 + 3x -10

x ≤  2    =   ( -inf , 2 ] 

https://web2.0calc.com/questions/counting_37747

(1 + i) (2 - i) (3 + i)

( 2 + 2i - 1i - i^2) (3 + i)

(2 + i  - (-1) ) (3 + i)

( 3 + i)  (3 + i)

9 + 3i + 3i  + i^2

9 + 6i  - 1

8 + 6i

https://web2.0calc.com/questions/simplify_39168